I'm trying to rename a number of files stored within subdirectories by removing the last four characters in their basename. I normally use glob.glob() to locate and rename files in one directory using:
import glob, os
for file in glob.glob("C:/Users/username/Desktop/Original data/" + "*.*"):
pieces = list(os.path.splitext(file))
pieces[0] = pieces[0][:-4]
newFile = "".join(pieces)
os.rename(file,newFile)
But now I want to repeat the above in all subdirectories. I tried using os.walk():
import os
for subdir, dirs, files in os.walk("C:/Users/username/Desktop/Original data/"):
for file in files:
pieces = list(os.path.splitext(file))
pieces[0] = pieces[0][:-4]
newFile = "".join(pieces)
# print "Original filename: " + file, " || New filename: " + newFile
os.rename(file,newFile)
The print statement correctly prints the original and the new filenames that I am looking for but os.rename(file,newFile) returns the following error:
Traceback (most recent call last):
File "<input>", line 7, in <module>
WindowsError: [Error 2] The system cannot find the file specified
How could I resolve this?
You have to pass the full path of the file to os.rename. First item of the tuple returned by os.walk is the current path so just use os.path.join to combine it with file name:
import os
for path, dirs, files in os.walk("./data"):
for file in files:
pieces = list(os.path.splitext(file))
pieces[0] = pieces[0][:-4]
newFile = "".join(pieces)
os.rename(os.path.join(path, file), os.path.join(path, newFile))
Related
I was trying to iterate over the files in a directory like this:
import os
path = r'E:/somedir'
for filename in os.listdir(path):
f = open(filename, 'r')
... # process the file
But Python was throwing FileNotFoundError even though the file exists:
Traceback (most recent call last):
File "E:/ADMTM/TestT.py", line 6, in <module>
f = open(filename, 'r')
FileNotFoundError: [Errno 2] No such file or directory: 'foo.txt'
So what is wrong here?
It is because os.listdir does not return the full path to the file, only the filename part; that is 'foo.txt', when open would want 'E:/somedir/foo.txt' because the file does not exist in the current directory.
Use os.path.join to prepend the directory to your filename:
path = r'E:/somedir'
for filename in os.listdir(path):
with open(os.path.join(path, filename)) as f:
... # process the file
(Also, you are not closing the file; the with block will take care of it automatically).
os.listdir(directory) returns a list of file names in directory. So unless directory is your current working directory, you need to join those file names with the actual directory to get a proper absolute path:
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
f = open(filepath,'r')
raw = f.read()
# ...
Here's an alternative solution using pathlib.Path.iterdir, which yields the full paths instead, removing the need to join paths:
from pathlib import Path
path = Path(r'E:/somedir')
for filename in path.iterdir():
with filename.open() as f:
... # process the file
I was trying to iterate over the files in a directory like this:
import os
path = r'E:/somedir'
for filename in os.listdir(path):
f = open(filename, 'r')
... # process the file
But Python was throwing FileNotFoundError even though the file exists:
Traceback (most recent call last):
File "E:/ADMTM/TestT.py", line 6, in <module>
f = open(filename, 'r')
FileNotFoundError: [Errno 2] No such file or directory: 'foo.txt'
So what is wrong here?
It is because os.listdir does not return the full path to the file, only the filename part; that is 'foo.txt', when open would want 'E:/somedir/foo.txt' because the file does not exist in the current directory.
Use os.path.join to prepend the directory to your filename:
path = r'E:/somedir'
for filename in os.listdir(path):
with open(os.path.join(path, filename)) as f:
... # process the file
(Also, you are not closing the file; the with block will take care of it automatically).
os.listdir(directory) returns a list of file names in directory. So unless directory is your current working directory, you need to join those file names with the actual directory to get a proper absolute path:
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
f = open(filepath,'r')
raw = f.read()
# ...
Here's an alternative solution using pathlib.Path.iterdir, which yields the full paths instead, removing the need to join paths:
from pathlib import Path
path = Path(r'E:/somedir')
for filename in path.iterdir():
with filename.open() as f:
... # process the file
I was trying to iterate over the files in a directory like this:
import os
path = r'E:/somedir'
for filename in os.listdir(path):
f = open(filename, 'r')
... # process the file
But Python was throwing FileNotFoundError even though the file exists:
Traceback (most recent call last):
File "E:/ADMTM/TestT.py", line 6, in <module>
f = open(filename, 'r')
FileNotFoundError: [Errno 2] No such file or directory: 'foo.txt'
So what is wrong here?
It is because os.listdir does not return the full path to the file, only the filename part; that is 'foo.txt', when open would want 'E:/somedir/foo.txt' because the file does not exist in the current directory.
Use os.path.join to prepend the directory to your filename:
path = r'E:/somedir'
for filename in os.listdir(path):
with open(os.path.join(path, filename)) as f:
... # process the file
(Also, you are not closing the file; the with block will take care of it automatically).
os.listdir(directory) returns a list of file names in directory. So unless directory is your current working directory, you need to join those file names with the actual directory to get a proper absolute path:
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
f = open(filepath,'r')
raw = f.read()
# ...
Here's an alternative solution using pathlib.Path.iterdir, which yields the full paths instead, removing the need to join paths:
from pathlib import Path
path = Path(r'E:/somedir')
for filename in path.iterdir():
with filename.open() as f:
... # process the file
I am using python to rename files which exist as binary files but in actual are images. So I need to rename them into .jpg format. I am using os.rename() but getting the error:
Traceback (most recent call last):
File "addext.py", line 8, in <module>
os.rename(filename, filename + '.jpg')
OSError: [Errno 2] No such file or directory
Here's my code.
import os
for filename in os.listdir('/home/gpuuser/Aditya_Nigam/lum2/'):
# print(filename + '.jpg')
# k = str(filename)
# print k
# k = filename + '.jpg'
os.rename(filename, filename + '.jpg')
print('Done')
os.listdir only return a list of filenames without their absolute paths, and os.rename will attempt to lookup a filename from the current directory unless given an absolute path. Basically, the code as-is will only work when executed in the same directory as the one called by os.listdir.
Consider doing the following:
import os
from os.path import join
path = '/home/gpuuser/Aditya_Nigam/lum2/'
for filename in os.listdir(path):
os.rename(join(path, filename), join(path, filename) + '.jpg')
The os.path.join method will safely join the path with the filenames together in a platform agnostic manner.
I'm trying to do a couple things here with the script below (it is incomplete). The first thing is to loop through some subdirectories. I was able to do that successfully. The second thing was to open a specific file (it is the same name in each subdirectory) and find the minimum and maximum value in each column EXCEPT the first.
Right now I'm stuck on finding the max value in a single column because the files I'm reading have two rows which I want to ignore. Unfortunately, I'm getting the following error when attempting to run the code:
Traceback (most recent call last):
File "test_script.py", line 22, in <module>
with open(file) as f:
IOError: [Errno 2] No such file or directory: 'tc.out'
Here is the current state of my code:
import scipy as sp
import os
rootdir = 'mydir'; #mydir has been changed from the actual directory path
data = []
for root, dirs, files in os.walk(rootdir):
for file in files:
if file == "tc.out":
with open(file) as f:
for line in itertools.islice(f,3,None):
for line in file:
fields = line.split()
rowdata = map(float, fields)
data.extend(rowdata)
print 'Maximum: ', max(data)
To open a file you need to specify full path. You need to change the line
with open(file) as f:
to
with open(os.path.join(root, file)) as f:
When you write open(file), Python is trying to find the the file tc.out in the directory where you started the interpreter from. You should use the full path to that file in open:
with open(os.path.join(root, file)) as f:
Let me illustrate with an example:
I have a file named 'somefile.txt' in the directory /tmp/sto/deep/ (this is a Unix system, so I use forward slashes). And then I have this simple script which resides in the directory /tmp:
oliver#armstrong:/tmp$ cat myscript.py
import os
rootdir = '/tmp'
for root, dirs, files in os.walk(rootdir):
for fname in files:
if fname == 'somefile.txt':
with open(os.path.join(root, fname)) as f:
print('Filename: %s' % fname)
print('directory: %s' % root)
print(f.read())
When I execute this script from the /tmp directory, you'll see that fname is just the filename, the path leading to it is ommitted. That's why you need to join it with the first returned argument from os.walk.
oliver#armstrong:/tmp$ python myscript.py
Filename: somefile.txt
directory: /tmp/sto/deep
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